- [Instructor] Hey folks.
In this video I wanna show you something pretty interesting
about these Lagrange multipliers that we've been studying.
So the first portion I'm just gonna kind of get this set up,
which is a lot of review from what we've seen already,
but I think you're really gonna like where this is going
in the end.
So one of the examples I showed, and I think this is a
pretty nice prototypical example
for constrained optimization problems,
is that you're running a company
and you have some kind of revenue function
that's dependent on various choices
you make in running the company,
and I think I said the number of hours of labor
you employ and the number of tons of steel you use,
you know, if you were manufacturing something metallic.
And, you know, this might be modeled as some multivariable
function of h and s, right now we don't really care
about the specifics.
And you're trying to maximize this, right,
that's kind of the whole point of this unit
that we've been doing, is that you're trying to maximize
some function, but you have a constraint.
This is the real world, you can't just spend infinite money,
you have some kind of budget, some sort of amount of money
you spend as a function of those same choices you make,
the hours of labor you employ, the tons of steel you use.
And this, again, it's gonna equal some
multivariable function that tells you, you know,
how much money you spend for a given amount of hours
and given number of tons of steel.
And you set this equal to some constant,
this tells you the amount of money you're willing to spend.
And our goal has been to maximize some function,
subject to a constraint like this.
And the mental model you have in mind
is that you're looking in the input space,
inside the x-y plane,
or I guess really, it's the h-s plane in this case, right.
Your inputs are h and s, and points in this plane
tell you possible choices you can make
for hours of labor and tons of steel.
And you think of this budget
as some kind of curve in that plane, right.
All the sets of h and s that equal $10,000
is gonna give you some kind of curve.
And the core value we care about is that,
when you maximize this revenue,
you know, when you set it equal to a constant
I'm gonna call M star, that's like the maximum possible
revenue, that's gonna give you a contour
that's just barely tangent
to the constraint curve.
And if that seems unfamiliar,
definitely take a look at the videos preceding this one.
But, just to kind of continue the review,
this gave us the really nice property
that you look at the gradient vector
for the thing you're trying to maximize, R,
and that's gonna be proportional to the gradient vector
for the constraint function, for this B, so gradient of B.
And this is because gradients are perpendicular
to contour lines.
Again, this should feel mostly like review at this point.
So the core idea was that we take this gradient of R,
and then make it proportional, with some kind of
proportionality constant lambda,
to the gradient of B, to the gradient
of the constraint function.
And up till this point, this value lambda
has been wholly uninteresting.
It's just been a proportionality constant, right,
because we couldn't guarantee that the gradient of R
is equal to the gradient of B.
All we care about is that they're pointing
in the same direction.
So we just had this constant sitting here,
and all we really said is make sure it's not zero.
But here, we're gonna get to where this little guy
So, if you'll remember, in the last video,
I introduced this function called the Lagrangian,
And it takes in multiple inputs,
they'll be the same inputs that you have
for your budget function and your revenue function,
or more generally, the constraint
and the thing you're trying to maximize.
It takes in those same variables,
but, also, as another one of its inputs,
it takes in lambda.
So, it is a higher-dimensional function
than both of these two, because we've got this extra lambda.
And the way it's defined looks a little strange at first,
it just seems kind of like this random
hodgepodge of functions that we're putting together.
But, last time, I kind of walked through
why this makes sense.
You take the thing you're trying to maximize,
and you subtract off this lambda,
multiplied by the constraint function,
which is B of those inputs,
minus, and then whatever this constant is here, right.
I'm gonna give it a name, I'm gonna call this constant
So maybe we're thinking of it as $10,000,
but it's whatever your actual budget is.
So we think of that, and I'm just gonna
emphasize here that that's a constant, right,
that this b is being treated as a constant right now.
You know, we're thinking of h and s and lambda
all as these variables, and this gives us
some multivariable function.
And if you'll remember from the last video,
the reason for defining this function
is it gives us a really nice compact way to solve
the constraint optimization problem.
You set the gradient of L equal to zero,
or really the zero vector, right, it'll be a vector
with three components here.
And when you do that, you'll find some solution, right,
you'll find some solution, which I'll call h star,
s star, and lambda, here I'll give it that green
lambda color, lambda star.
You'll find some value that, when you input this
into the function, the gradient will equal zero.
And, of course, you might find multiple of these, right,
you might find multiple solutions to this problem,
but what you do is, for each one of them you're gonna take
a look at h star and s star, then you're gonna plug those
into the revenue function, or the thing that you're trying
And, typically, you only get a handful, you get a number,
then you can actually plug each one of them into
the revenue function, and you'll just check which one
of them makes this function the highest.
And whatever the highest value this function can achieve,
that is M star, that is the maximum possible revenue,
subject to this budget.
But it's interesting that when you solve this, you get some
specific value of lambda, right, there's a specific
lambda star that will be associated with the solution.
And, like I said, this turns out not to just be
some dummy variable.
It's gonna carry information about how much we can increase
the revenue if we increase that budget.
And, here, let me show you what I mean.
So we've got this M star, and I'll just write it again,
M star here.
And what that equals, I'm saying that's the maximum
So that's gonna be the revenue when you evaluate it at
h star and s star.
And h star and s star, they are whatever the solution
to this gradient of the Lagrangian equals zero equation is.
You set this multivariable function equal to the
zero vector, you solve when each of its partial derivatives
equal zero, and you'll get some kind of solution.
So when you plug that solution in the revenue,
that gives you the maximum possible revenue.
But what we could do is consider this as a function
of the budget.
Now, this is the kind of thing that looks a little bit wacky
if you're just looking at the formulas.
But if you actually think about what it means
in this context of kind of a revenue and a budget,
I think it's actually pretty sensible,
where, really, if we consider this b no longer to be
constant but something that you could change, right,
you're wondering, well what if I had a $20,000 budget,
or what if I had a $15,000 budget?
You wanna ask the question, what happens as you change
Well, the maximizing value, h star and s star,
each one of those guys is gonna depend on b, right.
As you change what this constant is, it's gonna change
the values at which the gradient of the Lagrangian
So, I'm gonna rewrite this function as
the revenue evaluated at h star and s star,
but now I'm gonna consider that h star and s star
each as functions of b, right,
because they depend on it in some way.
As you change b, it changes the solution to this problem
It's very implicit and it's kind of hard to think about.
It's hard to think, okay, as I change this b, how much
does that change h star.
Well that depends on what the, you know,
what the definition of R was and everything there.
But, in principle, in this context, I think it's quite
You have a maximum possible revenue,
and that depends on what your budget is.
So, what turns out to be a beautiful, absolutely beautiful
is that this lambda star
is equal to the derivative of M star,
the derivative of this maximum possible revenue
with respect to b, with respect to the budget.
And let me just show you what that actually means, right.
So if, for example, let's say you did
all of your calculations and it turned out that lambda star
was equal to 2.3.
You know, previously that just seemed like some dummy number
that you ignore, and you just look at whatever the
associated values here are.
But if you plug this in the computer and you see lambda star
equals 2.3, what that means is, for a tiny change in budget,
like let's say your budget increases from 10,000
to 10,001, it goes up to $10,001,
you increase your budget by just a little bit, a little db.
Then the ratio of the change in the maximizing revenue
to that db is about 2.3.
So what that would mean is, increasing your budget
by a dollar is gonna increase M star,
over here it would mean that M star increases
by about, you know, $2.30
for every dollar that you increase your budget.
And that's information you'd wanna know, right?
If you see that this lambda star is a number
bigger than one, you'd say, hey, maybe we should increase
We increase it from $10,000 to 10,001 and we're making
So, maybe, as long as lambda star is greater than one,
you should keep doing whatever it takes to increase
So this fact is quite surprising, I think,
and it seems like it totally comes out of nowhere.
So what I'm gonna do in the next video is prove this to you,
is prove why this is true, why this lambda star value
happens to be the rate of change for the maximum value
of the thing we're trying to maximize
with respect to this constant, with respect to whatever
constant you set your constraint function equal to.
For right now, though, I just want you to kind of
try to sit back and digest what this means in the context
of this specific economic example.
And, even if you never looked into the proof
and never understood it there,
I think this is an interesting and even useful
tidbit of knowledge to have about Lagrange multipliers.
So with that, I'll see you in the next video.